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dx-1-cos-x-1-sin-x-




Question Number 167586 by cortano1 last updated on 20/Mar/22
        λ=∫ (dx/( (√(1+cos x))+(√(1+sin x)))) =?
λ=dx1+cosx+1+sinx=?
Commented by MJS_new last updated on 20/Mar/22
1^(st)  step t=tan (x/2)  2^(nd)  step u=t+(√(t^2 +1))  I′ve got no time right now...
1ststept=tanx22ndstepu=t+t2+1Ivegotnotimerightnow
Answered by greogoury55 last updated on 20/Mar/22
  ∫(((√(1+cos x))−(√(1+sin x)))/(cos x−sin x)) dx   = ∫ ((√(2cos^2 ((x/2))))/(cos x−sin x)) dx−∫ ((√((cos (1/2)x+sin (1/2)x)^2 ))/(cos x−sin x)) dx  = ∫ (((√2) cos (x/2))/(cos x−sin x)) dx−∫ ((cos (x/2)+sin (x/2))/(cos x−sin x)) dx
1+cosx1+sinxcosxsinxdx=2cos2(x2)cosxsinxdx(cos12x+sin12x)2cosxsinxdx=2cosx2cosxsinxdxcosx2+sinx2cosxsinxdx
Answered by MJS_new last updated on 20/Mar/22
∫(dx/( (√(1+cos x))+(√(1+sin x))))=       [t=tan (x/2) → dx=((2dt)/(t^2 +1))]  =2∫(dt/((t+1+(√2))(√(t^2 +1))))=       [u=t+(√(t^2 +1)) → dt=((√(t^2 +1))/u)du]  =4∫(du/(u^2 +2(1+(√2))u−1))=  =(√(2−(√2)))ln ((u+1+(√2)−(√(4+2(√2))))/(u+1+(√2)+(√(4+2(√2))))) =...
dx1+cosx+1+sinx=[t=tanx2dx=2dtt2+1]=2dt(t+1+2)t2+1=[u=t+t2+1dt=t2+1udu]=4duu2+2(1+2)u1==22lnu+1+24+22u+1+2+4+22=
Commented by peter frank last updated on 21/Mar/22
thank you
thankyou

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